By Arkadii M. Slinko

Arkadii Slinko, now on the college of Auckland, used to be one of many top figures of the USSR Mathematical Olympiad Committee over the last years ahead of democratisation. This booklet brings jointly the issues and options of the final 4 years of the All-Union arithmetic Olympiads. not just are the issues and suggestions hugely expository however the publication is worthy analyzing on my own for the interesting background of arithmetic competitions to be present in the advent.

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**Additional resources for USSR mathematical olympiads 1989-1992**

**Example text**

17 18 CHAPTER 1. 35. Why is it not possible to easily extend Fisk’s proof above to the case of polygons with holes? 36. 14, derive an upper bound on the number of guards needed to cover a polygon with h holes and n total vertices. ) When all edges of the polygon meet at right angles (an orthogonal polygon), fewer guards are needed, as established by Jeff Kahn, Maria Klawe, and Daniel Kleitman in 1980. In contrast, covering the exterior rather than the interior of a polygon requires (in general) more guards, established by Joseph O’Rourke and Derick Wood in 1983.

F (qv) = q f (v) for all q ∈ Q and v ∈ R ; 3. f (π) = 0 . We call any such function a d-function (d for dihedral). For instance, for any d-function f , we see that f 5π 2 = 5 5 · f (π) = · 0 = 0. 2 2 Similarly, f maps any rational multiple of π to 0. We define a rational angle as an angle that is a rational multiple of π, and an irrational angle as one that is not. For an edge e of a polyhedron, let l(e) denote the length of e and let φ(e) denote the dihedral angle of e. For any choice of d-function f , Dehn’s idea is to associate the value l(e) · f (φ(e)) to each edge e, which he called its mass.

Since this is equal to l(e) · f (φ(e)), the masses add up in the required manner. 2. 28(b). In this case, the sum of the masses becomes l(e) · f (φ(e)) = l(e) · f (π) = 0. So a new edge created from a dissection that appears in the interior of a face of P has no mass. 3. 28(c). By a similar argument as before, l(e) · f (2π) = 0, again contributing no new mass. Thus the mass sum under the dissection depends only on the edges of P. As each edge e is covered exactly once by dissection edges, whose lengths sum to l(e), the mass sum for any dissection is exactly the same as the mass sum for the original P.